Math Problem Statement

A sequence is defined by the recurrence relation $a_n = 11a_{n-1} + 3n$ for $n \geq 1$, with initial condition $a_0 = 1$. Find the smallest positive integer $n$ such that the sum of the first $n$ terms of the sequence is divisible by 11.

Solution

To solve the problem, let’s break it down step-by-step. We need to find the smallest positive integer nn such that the sum of the first nn terms of the sequence defined by the recurrence relation

an=11an1+3n,a0=1a_n = 11a_{n-1} + 3n, \quad a_0 = 1

is divisible by 11.


Step 1: Calculate a few terms of the sequence ana_n

We start with the initial condition a0=1a_0 = 1 and use the recurrence relation to compute the next few terms to understand the sequence better.

  • For n=1n = 1: a1=11a0+3(1)=111+3=14a_1 = 11a_0 + 3(1) = 11 \cdot 1 + 3 = 14

  • For n=2n = 2: a2=11a1+3(2)=1114+6=154+6=160a_2 = 11a_1 + 3(2) = 11 \cdot 14 + 6 = 154 + 6 = 160

  • For n=3n = 3: a3=11a2+3(3)=11160+9=1760+9=1769a_3 = 11a_2 + 3(3) = 11 \cdot 160 + 9 = 1760 + 9 = 1769

So, the first few terms are: a0=1,a1=14,a2=160,a3=1769,a_0 = 1, \quad a_1 = 14, \quad a_2 = 160, \quad a_3 = 1769, \quad \dots


Step 2: Define the Sum SnS_n

We are asked to find the smallest nn such that the sum of the first nn terms is divisible by 11. Let

Sn=a0+a1+a2++an.S_n = a_0 + a_1 + a_2 + \dots + a_n.

We need to find the smallest positive nn such that Sn0(mod11)S_n \equiv 0 \pmod{11}.


Step 3: Compute the Partial Sums

Let’s compute the partial sums step-by-step and reduce them modulo 11 to see when the divisibility condition is satisfied.

  • S1=a0+a1=1+14=15S_1 = a_0 + a_1 = 1 + 14 = 15
    154(mod11).15 \equiv 4 \pmod{11}.

  • S2=S1+a2=15+160=175S_2 = S_1 + a_2 = 15 + 160 = 175
    17510(mod11).175 \equiv 10 \pmod{11}.

  • S3=S2+a3=175+1769=1944S_3 = S_2 + a_3 = 175 + 1769 = 1944
    19447(mod11).1944 \equiv 7 \pmod{11}.

Let’s continue further.

  • S4=S3+a4S_4 = S_3 + a_4 (we need a4a_4): a4=11a3+3(4)=111769+12=19459+12=19471.a_4 = 11a_3 + 3(4) = 11 \cdot 1769 + 12 = 19459 + 12 = 19471. Thus: S4=1944+19471=21415.S_4 = 1944 + 19471 = 21415. 214153(mod11).21415 \equiv 3 \pmod{11}.

  • S5=S4+a5S_5 = S_4 + a_5 (we need a5a_5): a5=11a4+3(5)=1119471+15=214181+15=214196.a_5 = 11a_4 + 3(5) = 11 \cdot 19471 + 15 = 214181 + 15 = 214196. Thus: S5=21415+214196=235611.S_5 = 21415 + 214196 = 235611. 2356110(mod11).235611 \equiv 0 \pmod{11}.


Step 4: Conclusion

The smallest positive integer nn such that the sum of the first nn terms is divisible by 11 is:

5.\boxed{5}.


Would you like further details or clarifications?


Here are 5 related questions for deeper exploration:

  1. How can we find a closed formula for the sequence ana_n?
  2. What are the properties of sequences defined by linear recurrence relations?
  3. How can modular arithmetic be used more systematically in recurrence problems?
  4. Can generating functions help solve such recurrence relations?
  5. What is the asymptotic behavior of this sequence ana_n as nn \to \infty?

Tip: When working with recurrences, computing a few initial terms often gives crucial insights into patterns and helps with verification.

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Math Problem Analysis

Mathematical Concepts

Recurrence Relations
Modular Arithmetic
Summation of Sequences

Formulas

Recurrence relation: a_n = 11a_{n-1} + 3n
Sum of the sequence: S_n = a_0 + a_1 + ... + a_n
Modulus operation: S_n ≡ 0 (mod 11)

Theorems

Modular Arithmetic
Linear Recurrence Relations

Suitable Grade Level

Grades 10-12